Question

Let $\overrightarrow{a},\overrightarrow{b}$ be two vectors such that $\left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=4,\overrightarrow{a}.\overrightarrow{b}=2$. If $\overrightarrow{c}=2(\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b}$, then the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is

• A. $\frac{2\pi }{3}$
• B. $\frac{5\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

$\frac{5\pi }{6}$

${|\overline{a}\ast \overline{b}|}^2={\left|\overline{a}\right|}^2{\left|\overline{b}\right|}^2-(\overline{a}.\overline{b}{)}^2$
$=16-4$
$=12$
Now ${\left|\overline{c}\right|}^2=4{|\overline{a}\ast \overline{b}|}^2+9{\left|\overline{b}\right|}^2-6\overline{a}\ast \overline{b}.\overline{b}$
$=48+144-0=192$
$\Rightarrow \left|\overline{c}\right|=8\sqrt{3}$
Also $\overline{b}.\overline{c}=-3{\left|\overline{b}\right|}^2=-48$
i.e., $\left|\overline{b}\right|\left|\overline{c}\right|cos\theta =-48$
$\Rightarrow 32\sqrt{3}cos\theta =-48$
$\Rightarrow cos\theta =-\frac{\sqrt{3}}{2}$
$\Rightarrow \theta =\frac{5\pi }{6}$